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posted by LSAT Thursday, June 2, 2011 at 7:58pm.

Find all solutions of equation on interval [0,2pi] 1=cot^2x + cscx

Multiply both sides by sin^2(x). sin^2(x)=cos^2(x)+sin(x) sin^2(x)=1-sin^2(x)+sin(x) 2sin^2(x)-sin(x)-1=0 (2sin(x)+1)(sin(x)-1)=0 sin(x)=-1/2, x=7pi/6, x=11pi/6 or sin(x)=1, x=pi/2

Thank you so much!

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